标签：ever 描述 new you self wap example pair efi
题目描述:
Given a linked list, swap every two adjacent nodes and return its head.
For example,
Given1->2->3->4, you should return the list as 2->1->4->3.
Your algorithm should use only constant space.
You may not modify the values in the list, only nodes itself can be changed.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
解法1：
递归方法
代码:
class Solution {
public:
ListNode *swapPairs(ListNode *head) {
if(head==nullptr || head->next==nullptr)
return head;
ListNode *next = swapPairs(head->next->next);
ListNode *newHead = head->next;
newHead->next = head;
head->next = next;
return newHead;
}
};
解法2：
直接方法
代码：
class Solution {
public:
ListNode *swapPairs(ListNode *head) {
if(head==nullptr || head->next==nullptr)
return head;
ListNode dummy = ListNode(-1);
ListNode *last = &dummy;
ListNode *cur = head;
while(cur!=nullptr && cur->next!=nullptr)
{
ListNode *next = cur->next->next;
last->next = cur->next;
last->next->next = cur;
last->next->next->next = nullptr;
last = last->next->next;
cur = next;
}
if(cur!=nullptr)
last->next = cur;
return dummy.next;
}
};
标签：ever 描述 new you self wap example pair efi
原文地址：https://www.cnblogs.com/zjuhaohaoxuexi/p/11788096.html
